Question

An artificial satellite of mass m is moving in a circular orbit at a height equal to the radius R of the earth. Suddenly, due to an internal explosion, the satellite breaks into two parts of equal pieces. One part of the satellite stops just after the explosion. The increase in the mechanical energy of the system due to explosion will be (Given acceleration due to gravity is g)
A. $mgR$
B. $\dfrac{{mgR}}{2}$
C. $\dfrac{{mgR}}{4}$
D. $\dfrac{{3mgR}}{4}$

Answer 1

Hint:Mechanical energy is the total sum of the kinetic and potential energies of a system. Thus, we have to calculate the kinetic energy in this case. To calculate the kinetic energy, we need to find the velocity as the relation between the quantities are –
Kinetic energy, $K = \dfrac{1}{2}m{v^2}$
where m = mass and v = velocity.
Also, the formula for acceleration due to gravity is given by –
$g = \dfrac{{GM}}{{{R^2}}}$
where G = gravitational constant, M = mass of earth and R = radius of earth.

Complete step-by-step answer:
Consider a satellite of mass m moving around the Earth’s surface at a height, h = R which is the radius of Earth.
The velocity of the satellite at this height is given by –
${v_0} = \sqrt {\dfrac{{GM}}{{R + h}}} $
where G = gravitational constant and M = mass of earth
Substituting the height, h = R
${v_0} = \sqrt {\dfrac{{GM}}{{R + R}}} = \sqrt {\dfrac{{GM}}{{2R}}} $
Now, the satellite explodes and splits into two equal pieces. Velocity of the first piece is 0 and the velocity of the second piece is v.

To understand the explosion, we need to apply the principle of conservation of momentum.
Conservation of momentum states that the total momentum before and after the event remains constant.
Hence,
Total momentum before the explosion = Total momentum after the explosion
$m{v_0} = \dfrac{m}{2} \times 0 + \dfrac{m}{2} \times v$
Hence, we get the relation –
$v = 2{v_0}$
The total increase in the mechanical energy is equal to the change in the potential energy plus the change in total kinetic energy.
$\Delta E = \Delta P + \Delta K$
Since, there is no change in the orbit of the satellite after the explosion, the gravitational potential energy remains the same. Hence, $\Delta P = 0$
Thus, $\Delta E = \Delta K$
Change in Kinetic energy,
$\Delta K = \dfrac{1}{2}\left( {\dfrac{m}{2}} \right){v^2} - \dfrac{1}{2}mv_0^2$
Substituting for v,
$\Delta K = \dfrac{1}{2} \times 4 \times \left( {\dfrac{m}{2}} \right) \times v_0^2 - \dfrac{1}{2}mv_0^2$
$ \Rightarrow \Delta K = mv_0^2 - \dfrac{1}{2}mv_0^2 = \left( {1 - \dfrac{1}{2}} \right)mv_0^2 = \dfrac{1}{2}mv_0^2$
Hence,
$\Delta E = \dfrac{1}{2}mv_0^2$
Substituting the value of ${v_0} = \sqrt {\dfrac{{GM}}{{2R}}} $ in the above equation,
$\Delta E = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{{2R}}} } \right)^2}$
$ \Rightarrow \Delta E = \dfrac{1}{2}m \times \dfrac{{GM}}{{2R}}$
The acceleration due to gravity of earth, $g = \dfrac{{GM}}{{{R^2}}}$
$ \Rightarrow gR = \dfrac{{GM}}{R}$
Substituting, we get –
$\Delta E = \dfrac{1}{2}m \times \dfrac{1}{2}gR = \dfrac{{mgR}}{4}$

Hence, the correct option is Option C.

Note:Derivation of orbital velocity of satellite.
When a satellite is rotating about the Earth, the centripetal force is provided by the Earth. Hence,
$\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
where r = radius of Earth, R + height of the satellite, h.
Solving,
${v^2} = \dfrac{{GM}}{r}$
$\therefore v = \sqrt {\dfrac{{GM}}{r}} $
where r = R + h